If the system of equations $x+y+z=2$,$2x+4y-z=6$,and $3x+2y+\lambda z=\mu$ has infinitely many solutions,then:

Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ be a $2 \times 2$ real matrix with $\det(A) = 1$. If the equation $\det(A - \lambda I_2) = 0$ has imaginary roots (where $I_2$ is the identity matrix of order $2$), then:

If $AX=B$,where $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 2 & 1\end{array}\right]$,$X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$,and $B=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$,then find the value of $2x+y-z$.

If the values of $x, y$ and $z$ which satisfy the equations $2x - 3y + 2z + 15 = 0$,$3x + y - z + 2 = 0$ and $x - 3y - 3z + 8 = 0$ simultaneously are $\alpha, \beta$ and $\gamma$ respectively,then:

The values of $\lambda$ and $\mu$ for which the system of equations $x+y+z=6, x+2y+3z=10, x+2y+\lambda z=\mu$ has infinitely many solutions are

If the system of equations $2x + py + 6z = 8$,$x + 2y + qz = 5$,and $x + y + 3z = 4$ has infinitely many solutions,then $p=$